what is the differential equation of (2x+y+1)dx + (2y+x+1)dy=0
YES
To find the differential equation of the given expression (2x+y+1)dx + (2y+x+1)dy = 0, we can start by rearranging the equation into a more standard form.
Begin by separating the variables of x and y.
(2x + y + 1)dx + (2y + x + 1)dy = 0
(2x + y + 1)dx = -(2y + x + 1)dy
Next, integrate both sides with respect to their respective variables.
∫ (2x + y + 1)dx = -∫ (2y + x + 1)dy
Integrating the left side with respect to x:
∫ (2x + y + 1)dx = ∫ 2xdx + ∫ ydx + ∫ 1dx
= x^2 + xy + x + C1, where C1 is the constant of integration for the x variable.
Integrating the right side with respect to y:
-∫ (2y + x + 1)dy = -∫ 2ydy - ∫ xdy - ∫ 1dy
= -y^2 - xy - y + C2, where C2 is the constant of integration for the y variable.
Now we have:
x^2 + xy + x + C1 = -y^2 - xy - y + C2
Rearrange this equation to isolate the variables on one side:
x^2 + xy + y^2 + x + y + C1 - C2 = 0
Finally, simplify the constant terms:
x^2 + xy + y^2 + x + y + C = 0,
where C = C1 - C2 is the new constant.
Therefore, the differential equation of (2x+y+1)dx + (2y+x+1)dy = 0 is x^2 + xy + y^2 + x + y + C = 0, where C is a constant.